# The conditional pdf of 3 iid random variables from an exponential distribution

Consider three iid random varaiables from an exponential distribution, $X_1,X_2,X_3 \sim$ Exponential$(1)$, each with a pdf $f(x) = e^{-x} , \space \text{where } x >0$. Determine the conditional pdf of the maximum of the sample given using the minimum of the sample, i.e. $f_{X_{(3)}|X_{(1)}} (y \space|\space x)$.

My process

For a continuous random variable the conditional pdf of a probability function of $X$ given $Y$ is $$f_{X|Y}(x,y) = \dfrac{f(x,y)}{f_Y(y)}$$ From what I can find online, to find the conditional pdf of $Y$ given $X$ we have to find $f_X(x)$, the marginal pdf of $X$ first, which I can do by integrating the joint pdf $f(X_1,X_2,X_3)$. How do I find the joint pdf of these three variables?

Well, the joint pdf of the three independent samples is the product of their marginal pdf, which you already know.   That is $f_X(z)=e^{-z}\mathbf 1_{x\geqslant 0}$, and also $F_X(z)=(1-e^{-z})\,\mathbf 1_{x\geqslant 0}$ is the CDF of the random variable.   So $f_{X_1,X_2,X_3}(u,v,w) = f_X(u)\,f_X(v)\,f_X(w)$

However, no ; you don't just integrate the joint pdf of the samples; that will find you the marginal of the first sample, not the least sample.   You want to find the First Order Statistic's pdf.

The favoured outcomes of this (zero measure) event are those consisting of: a selection of one from three samples to be at the value, with the other two samples greater than it.   Hence the probability density function of the first order statistic is:

$$f_{X_{(1)}}(x) = 3\, f_X(x)\,(1-F_X(x))^2$$

By similar reasoning the joint pdf of the minimum and maximum sample values is the density measure of the event where : we select two samples with these two values while the third lies somewhere between them.:

$$f_{X_{(1)},X_{(3)}}(x,y) = 3!\, f_X(x)\,f_{X}(y)\,(F_X(y)-F_X(x))$$

Hence $$f_{X_{(3)}\mid X_{(1)}}(y\mid x) ~=~\dfrac{2 f_X(y)\,(F_X(y)-F_X(x))}{(1-F_X(x))^2}$$

Substitute and simplify.

• In this case, what would $f_X(y)$ and $F_X(y)$ actually be? Silly question I know, but because of the three variables... – Rubicon Sep 6 '17 at 13:26
• $f_X(y)~=~e^{-y}\mathbf 1_{y\geqslant 0}$ and $F_X(y)~=~(1-e^{-y})\mathbf 1_{y\geqslant 0}$. Just a change of tokens. (iid:= independent and identically distributed) $f_X(y)$ is the probability density function for an arbitrary one of the three variables measured at $y$. @GBTV – Graham Kemp Sep 6 '17 at 13:44